Question

giúp em bài 1 dạng 3 vói em, em cần gấp lắm, em cảm ơn ạ

Answer 1

Bài 1:

1) \(\Rightarrow\left(x-2\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

2) \(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

3) \(\Rightarrow\left(4x-3\right)\left(7-12x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{7}{12}\end{matrix}\right.\)

4) \(\Rightarrow x^3+8-x^3+25x=-17\)

\(\Rightarrow25x=-25\Rightarrow x=-1\)

5) \(\Rightarrow\left(3x-2\right)\left(3x+2\right)-2\left(3x-2\right)^2=0\)

\(\Rightarrow\left(3x-2\right)\left(3x+2-6x+4\right)=0\)

\(\Rightarrow\left(3x-2\right)\left(-3x+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

Answer 2

Bài 3: 

c: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

d: \(x^3-7x-6\)

\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x-6\right)\)

\(=\left(x+1\right)\left(x-3\right)\left(x+2\right)\)